Poisson probability distribution Question....Stuck!

Hey would really, really appreciate some help with this question on poisson distribution, this is like the only thing on my stats module so far I cant seem to understand. Lol, it just won't seem to click.

Q:
Estimate the number of serious road accidents a company can expect during mining operations. They know the average number of accidents on this type of mining site is .5 and follows a poisson distribution.

Calculate the probability that
(i) less than 2 accidents on site A
(ii) more than 4 accidents on site A
(iii) 3 or more accidents in total on all 3 sites

So, is it something like this?
Less than 2 accidents, P(X<2) = P(0)+P(1)
I am confused
Thank you!

In (c) I assume we are talking about 3 indep sites all Poisson with mean .5.
The sum of indep Poisson's is a Poisson.

Hey thanks so much for help. Sorry I should have mentioned site A is independent of B & C as these two are same. Is there much work involved in the solutions for these? I have been trying to do a few examples from different sites but am a bit confused as some of them seem to write the formula a different way.
Many thanks.

This is just from an example and while I can see the process I don't know how to go about calculating (2.718)-2, so how do you go about it using a calculator?
Thanks

It's easy,first of all if you're using a scientific calculator say... Casio fx-991ES,
"e" is just a constant on it.
In other words, you just press it and it allows you to raise it to any power. In your case -2, because λ=2 and e^-λ right? So your calculator should show "e^( " when you press "Shift" then "In" once each.

Otherwise, you may key in the equivalent of "e" i.e 2.71828 and raise it to power -2 on your calculator. This should show 2.71828^(-2).